[tex]\bf sin^2(\theta)+cos^2(\theta)=1\\\\
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y=\cfrac{1+sin(x)}{1-cos(x)}\impliedby \textit{using the quotient rule}
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\cfrac{dy}{dx}=\cfrac{cos(x)[1-cos(x)]-[1+sin(x)]sin(x)}{[1-cos(x)]^2}
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\cfrac{dy}{dx}=\cfrac{cos(x)-cos^2(x)-sin(x)-sin^2(x)}{[1-cos(x)]^2}
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\cfrac{dy}{dx}=\cfrac{cos(x)-sin(x)-[cos^2(x)+sin^2(x)]}{[1-cos(x)]^2}
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\cfrac{dy}{dx}=\cfrac{cos(x)-sin(x)-\boxed{1}}{[1-cos(x)]^2}
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\cfrac{dy}{dx}=\cfrac{-sin(x)+cos(x)-1}{[1-cos(x)]^2}[/tex]
